Simplify the following expression and state the condition under which the simplification is valid. You can assume that $t \neq 0$. $z = \dfrac{2t - 4}{-10} \times \dfrac{8t}{8t(t - 2)} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ (2t - 4) \times 8t } { -10 \times 8t(t - 2) } $ $ z = \dfrac {8t \times 2(t - 2)} {-10 \times 8t(t - 2)} $ $ z = \dfrac{16t(t - 2)}{-80t(t - 2)} $ We can cancel the $t - 2$ so long as $t - 2 \neq 0$ Therefore $t \neq 2$ $z = \dfrac{16t \cancel{(t - 2})}{-80t \cancel{(t - 2)}} = -\dfrac{16t}{80t} = -\dfrac{1}{5} $